Question

Let y=f(x) be the given curve and $x=a$, $x=b$ be two ordinates then area bounded by the curve $y=f\left(x\right)$, the axis of x between the ordinates $x=a$ & $x=b$, is given by definite integral
${\int }_a^{b}ydx$ or ${\int }_a^{b}f\left(x\right)dx$ and the area bounded by the curve $x=f\left(y\right)$, the axis of y & two abscissae $y=c$ & $y=d$ is given by ${\int }_c^{d}xdy$ or ${\int }_c^{d}f\left(x\right)dy$. Again if we consider two curves $y=f\left(x\right)$, $y=g\left(x\right)$ where $f\left(x\right)\ge g\left(x\right)$ in the interval [a, b] where $x=a$ & $x=b$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
${\int }_a^b[f\left(x\right)-g\left(x\right)]dx$
On the basis of above information answer the following questions.

The area bounded by the curves $y=\sqrt{x}$, $x=2y+3$ and x-axis (in the first quadrant) is equal to (in square units)

• A. $9$
• B. $12$
• C. $6$
• D. $\frac{9}{2}$

Answer 1

Answer: (A)

$9$

Given curves are, $y=\sqrt{x}$ (i)
$y=\frac{x-3}{2}$ (ii)
and x-axis i.e. $y=0$ (iii)
Now equation (i) is $y^2=x$ is right handed parabola but with positive values of y the part of the curve lying above x-axis
Now solving (i) & (ii) we get $4x={(x-3)}^2$
$\Rightarrow$ $x^2-10x+9=0$ $\Rightarrow$ $(x-1)(x-9)=0$
$\Rightarrow$ $x=1,x=9$, rejecting $x=1$ as it gives $y=-1$
$\therefore$ $x=9,y=3$
$\therefore$ Required area $={\int }_0^9\sqrt{x}dx-{\int }_3^9\left(\frac{x-3}{2}\right)dx$
$=18-9=9$square units